# Inverse results in special relativity

• I
• Kairos
In summary, the conversation discusses the relationship between energy, mass, and frequency in a relativistic context. The equations show that as an object's speed increases, its mass increases and its apparent frequency decreases. However, there is no direct relationship between the mass and frequency of an object, and the use of the de Broglie relation for photons to explain this relationship is outdated and considered nonsense. To properly consider relativistic effects, one must use a relativistic form of quantum mechanics.
Kairos
The mass of an object moving at speed v increases such that $$\frac{m'}{m}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

and its apparent frequency decreases such that $$\frac{\nu'}{\nu}=\sqrt{1-\frac{v^{2}}{c^{2}}}$$

so $$\frac{\nu'}{\nu}=\frac{m}{m'}$$

but equating the energies $$h\nu= mc^{2}$$

gives the inverse relationship $$\frac{\nu'}{\nu}=\frac{m'}{m}$$
Has this question already been addressed? Otherwise, thank you in advance if someone can point out my mistake.

Kairos said:
The mass of an object moving at speed v increases such that $$\frac{m'}{m}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

In order to avoid answers like #2 you better repeat your question with energy instead of mass:

$$\frac{E}{E_0}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

$$h\nu= E$$

When you do that you might already see what your problem actually results from.

vanhees71
DrStupid said:
In order to avoid answers like #2 you better repeat your question with energy instead of mass:

$$\frac{E}{E_0}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

Thanks. Yes, I misspoke; but I still don't understand. This ratio of energies corresponds that of wavelengths, not of frequencies

$$\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{\lambda}{\lambda_0}=\frac{\nu_0}{\nu}$$

does it not?

Kairos said:
The mass of an object moving at speed v

Why do you say such an object has a frequency ##\nu##? Is it executing simple harmonic motion or something like that? If so, it doesn't have a constant speed.

Mister T said:
Why do you say such an object has a frequency ##\nu##? Is it executing simple harmonic motion or something like that? If so, it doesn't have a constant speed.

OK I suppose I see what you mean: the above relationship between energies does not hold for a harmonic resonator in a rocket at speed v?

To the extent that the de Broglie relation can be generalised to 4-momentum and 4-frequency, your relation between the frequencies is wrong.

Kairos said:
its apparent frequency decreases

Note that when I use apparent I mean I am measuring the whatever quantity associated to the rocket wrt ##S##. (stationary frame).

I assume that we are dealing with, let's say, a rocket with a flashing light on it getting further and further away from a stationary frame (if it were approaching it instead, the apparent frequency ##\nu## would increase; see Dopler effect: https://en.wikipedia.org/wiki/Relativistic_Doppler_effect).

The apparent wavelength will increase; let me explain why.

A time ##t## elapses between flashes, so during this time the light will have traveled a distance ##ct##. But, of course, during that time the spaceship has also traveled a distance: ##ut##. Then the apparent wavelength will be:

$$\lambda = (c + u)t$$

Note that ##t## is the time as measured by an observer in ##S## (it's the 'apparent' time). As moving clocks run slow:

$$t = \gamma \bar t$$

Where ##\bar t## is the time as measured by an observer in ##\bar S##.

Then the apparent frequency is:

$$\nu = \frac{c}{\lambda} = \frac{c}{(c + u)t} = \frac{c(\sqrt{1 - u^2/v^2})}{(c + u)\bar t} = \frac{\sqrt{c - u}}{\sqrt{c + u}}\bar \nu$$

Where ##\bar \nu## is the rest frequency (I have used ##\bar t = \frac{1}{\bar \nu}##).

Actually professor Michel van Biezen works out the same exercise but with an approaching spaceship:

Hope at this point it is more clear why the frequency decreases wrt ##S## frame.

Kairos said:
OK I suppose I see what you mean: the above relationship between energies does not hold for a harmonic resonator in a rocket at speed v?

That is not my point. My point is that you have an object of mass ##m## moving with speed ##v##. Such an object will indeed have a rest energy of ##mc^2##. But you lose me when you say its energy is equal to ##h\nu##. You don't tell us what ##\nu## is. It's a frequency, but the frequency of what?

##h\nu## is the energy of a photon, but a photon has zero mass, thus ##mc^2=0##.

JD_PM said:
I assume that we are dealing with, let's say, a rocket with a flashing light on it getting further and further away
That can't be the case. The mass ##m## of the rocket is not in any way related to the frequency ##\nu## of the flashing light. In other words, there is no relation between ##m## and ##\nu##, therefore expressions like ##mc^2=h\nu## can't be valid and are founded on nonsense.

It's true that I just focused on the frequency. Maybe a proton moving with a relativistic speed is a good example here.

But let's wait for clarification.

Mister T said:
That can't be the case. The mass ##m## of the rocket is not in any way related to the frequency ##\nu## of the flashing light. In other words, there is no relation between ##m## and ##\nu##, therefore expressions like ##mc^2=h\nu## can't be valid and are founded on nonsense.
Are you calling the de Broglie relation nonsense? I mean, seriously outdated, fine, but nonsense seems to be stretching it a bit far.

PeroK
##E = h \, \nu## applies to a photon, but the frequency shift for a photon is given by the relativistic doppler formula

$$\frac{f_s}{f_r} = \sqrt{\frac{1+\beta}{1-\beta}}$$

This is important to understand for the correct expression of how the energy and momentum of a photon transforms relativistically.

For non-photons, if you want to consider relativistic effects, you need to use a relativistic form of quantum mechanics. For instance, you could use the Klein-Gordon equations for a particle with no spin, or the Dirac equations for a particle with spin 1/2.

Wiki has <<this link>> on relativistic wave equations. I'm not really all that familiar with the topic, unfortunately, and I can't say how good it is. You could try the quantum forum if you need a better answer.

Apparently, though, one can still use the Schrodinger wave equation

$$i\,h\frac{\partial}{\partial t} \, \Psi = H \, \Psi$$

but the relativisitc Hamiltonian operator, H, is not the non-relativistic one involving ##\frac{\hbar^2}{2m}##. Basically, this form of the hamiltonian comes from ##p^2/2m##, where p is the momentum, and you can see that it's a classical expression, not a relativistic one.

Mister T said:
That is not my point. My point is that you have an object of mass ##m## moving with speed ##v##. Such an object will indeed have a rest energy of ##mc^2##. But you lose me when you say its energy is equal to ##h\nu##. You don't tell us what ##\nu## is. It's a frequency, but the frequency of what?

##h\nu## is the energy of a photon, but a photon has zero mass, thus ##mc^2=0##.

Understood. The form of Energy
##h\nu## is not generalisable to objects with a mass through de Broglie wavelength?

Kairos said:
Understood. The form of Energy
##h\nu## is not generalisable to objects with a mass through de Broglie wavelength?
As far as the de Broglie relation is applicable (with all the caveats of being outdated as per the above remarks), yes it is. The relativistic variant of the de Broglie relation is ##h N = P = mV##, where ##N## is the 4-frequency, ##P## the 4-momentum, ##m## the invariant mass, and ##V## the 4-velocity of the object. Note that I am using units where ##c = 1##.

Kairos said:
Thanks. Yes, I misspoke; but I still don't understand. This ratio of energies corresponds that of wavelengths, not of frequencies

$$\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{\lambda}{\lambda_0}=\frac{\nu_0}{\nu}$$

does it not?
It is unclear to me why you think that
$$\frac{\lambda}{\lambda_0} = \gamma(v).$$
This is not the case for a matter wave, you cannot apply length contraction on wavelengths.

Orodruin said:
you cannot apply length contraction on wavelengths.

Why not?

JD_PM said:
Why not?
Why would you? There is not a single thing about a matter wave that lines up with length contraction for the wavelength. The phase velocity is > c.

Orodruin said:
Why would you?

I asked because while working out post #8 I thought 'why not ##\lambda = \gamma (c + u)t##?; ##(c + u)t## has dimensions of length, and we know that moving objects get shortened along the direction of motion...'.

JD_PM said:
I asked because while working out post #8 I thought 'why not ##\lambda = \gamma (c + u)t##?; ##(c + u)t## has dimensions of length, and we know that moving objects get shortened along the direction of motion...'.
You really should stop thinking in terms of length contraction and time dilation if you cannot easily identify the cases where they are applicable and those where they are not. Matter waves have no ”rest frame”. Their wavelength when in the rest frame of the corresponding object is infinite.

SiennaTheGr8 and JD_PM
Orodruin said:
Are you calling the de Broglie relation nonsense?
No, of course not. I didn't realize that ##\nu## was a frequency associated with the de Broglie wavelength.

If that's the case, then the derivation presented in the OP is flawed because the energy ##E## of an object of mass ##m## moving with speed ##v## is not equal to ##h\nu##.

Kairos
Orodruin said:
You really should stop thinking in terms of length contraction and time dilation if you cannot easily identify the cases where they are applicable and those where they are not. Matter waves have no ”rest frame”. Their wavelength when in the rest frame of the corresponding object is infinite.

Indeed. I've seen experts casually make this mistake in books, chalking something up to length contraction when it doesn't apply. I can think of two examples off the top of my head, actually. If unsure, identify events and use the Lorentz transformation!

Going back to my original problem, by combining posts 3 and 4,

## \frac{E}{E_0}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{\lambda}{\lambda_0}=\frac{\nu_0}{\nu} ##

when the energies are rather supposed to be proportional to the frequencies through ## E= h\nu ##
Does it mean that the first identity holds for a mass but is inverted for a photon?

Kairos said:
Going back to my original problem, by combining posts 3 and 4,

## \frac{E}{E_0}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{\lambda}{\lambda_0}=\frac{\nu_0}{\nu} ##

when the energies are rather supposed to be proportional to the frequencies through ## E= h\nu ##
Does it mean that the first identity holds for a mass but is inverted for a photon?
No. Again, your relationship between the wavelengths is wrong. In addition, ##\lambda/\lambda_0 = \nu_0/\nu## only holds if the speed of the wave is independent of the reference frame, which is also not the case.

Orodruin said:
Why would you? There is not a single thing about a matter wave that lines up with length contraction for the wavelength. The phase velocity is > c.

I would argue that if one is using a relativistic formulation of quantum mechanics (and I would guess the Original Poster is most likely not doing this), then the wave function of a single particle should transform via the Lorentz transform.

As an aside, I'm not quite sure of the details of how a wavefunction of a multi-particle system would transform. But let's stick to single particle wavefunctions for the time being.

I would expect that that Lorentz contraction would be part of (but not the complete description of) how the wavefunction transforms.

It's possible that some confusion could arise from omitting the other features of the Lorentz transform other than length contraction (time dilation and the relativity of simultaneity, to be specific). I don't think that's the main problem here, though.

I suppose a more specific problem could be useful, one might imagine asking what the allowed energy states of a particle in a box were, and perversely insist on working the problem out in a frame of reference where the box was moving at relativistic velocities, rather than in the rest frame of the box.

The box would length contract in such a scenario, and so would the wavefunction of the particle, which would be bounded if one makes the usual assumption of the box being an infinitely deep potential well.

Of course it'd be wrong to just consider length contraction of the box, and ignore other aspects arising from the Lorentz transform, but I don't think that's necessarily the issue here.

The usual textbook treatment of QM, not being a relativistic formulation, would probably not give covariant answers for the problem of a particle in a relativistic box.

pervect said:
As an aside, I'm not quite sure of the details of how a wavefunction of a multi-particle system would transform. But let's stick to single particle wavefunctions for the time being.
Well, a priori you cannot do this unless you have a non-interacting theory. This is one of the beauties with QFT.

What I am talking about is essentially related to the wave solutions of the classical Klein-Gordon equation, which to some extent generalise the de Broglie relation when interpreted as one-particle states of the non-interacting theory.

Orodruin said:
Again, your relationship between the wavelengths is wrong.

OK, in fact I was just thinking in term of transverse Doppler

Orodruin said:
In addition, ##\lambda/\lambda_0 = \nu_0/\nu## only holds if the speed of the wave is independent of the reference frame, which is also not the case.

I assumed ##c=\lambda \nu ##

thank you

Kairos said:
I assumed c=λν

Huhh? Are you still talking about an object of mass ##m## moving with speed ##v##? If so, the speed is ##c## only when the mass is zero. Relations like ##E=E_o \frac{1}{\sqrt{1-(v/c)^2}}## are not valid for objects moving at speed ##c##.

Kairos
Now I think I understand. Thank you.

Orodruin said:
Well, a priori you cannot do this unless you have a non-interacting theory. This is one of the beauties with QFT.

What I am talking about is essentially related to the wave solutions of the classical Klein-Gordon equation, which to some extent generalise the de Broglie relation when interpreted as one-particle states of the non-interacting theory.

That's more or less what I was suggesting as well. If one used the Klein-Gordon equations, a single particle wavefunction would behave more-or-less as the Original Poster expected in terms of "Lorentz contraction".

The Dirac equations might also apply, if he was doing QM with particles that had half-integral spin.

Most likely the OP is using the de-Broglie relation withtout thinking about where it came from. And he is confused by the fact that the answers when he does this are not relativistic. I suggest that the answers aren't relativistic because the theory he is using isn't relativistic. If he re-visited the problem using a relativistic version of quantum theory, such as the Klein-Gordon equation, or some other relativistic formulation, I believe he'd solve his immediate problem. Most likely he'd find some new things to be puzzled about if he got that far, but no need to borrow confusion in advance.

But my point is that there is a relativistic analogue of the de Broglie relation. It just relates 4-frequency to 4-momentum.

## 1. What are inverse results in special relativity?

Inverse results in special relativity refer to the concept that the laws of physics are the same for all observers in uniform motion, regardless of their relative velocities. This means that if one observer sees a certain event happen in a certain way, another observer moving at a different velocity would also see the same event happening in a different way, but the laws of physics governing the event would still be the same.

## 2. How does the concept of time dilation relate to inverse results in special relativity?

Time dilation is a direct consequence of inverse results in special relativity. It states that time passes slower for objects in motion compared to objects at rest. This means that if two observers are moving at different velocities, they will measure different amounts of time passing for the same event. However, the laws of physics governing the event will remain the same for both observers.

## 3. Can inverse results in special relativity be observed in everyday life?

Yes, inverse results in special relativity can be observed in everyday life. For example, the Global Positioning System (GPS) works by taking into account the time dilation effect caused by the high velocities of the satellites in orbit. If this effect was not taken into account, the GPS system would be inaccurate by several kilometers.

## 4. How does the principle of relativity play a role in inverse results in special relativity?

The principle of relativity states that the laws of physics are the same for all observers in uniform motion. This principle is the foundation of inverse results in special relativity, as it allows for the concept that the laws of physics remain the same for all observers, regardless of their relative velocities.

## 5. Are there any practical applications of inverse results in special relativity?

Yes, there are several practical applications of inverse results in special relativity. Apart from the GPS system mentioned earlier, this concept is also used in particle accelerators, where particles are accelerated to high velocities and their behavior is observed by different observers. It is also important in the study of high-speed spacecraft and in understanding the behavior of objects moving at near-light speeds.

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